The total angular momentum of the system about point B is 
Angular momentum, also known as moment of momentum or rotational momentum, is the rotating counterpart of linear momentum.
A rigid object's angular momentum is defined as the product of its moment of inertia and its angular velocity. If there is no external torque on the object, it is analogous to linear momentum and is subject to the fundamental constraints of the conservation of angular momentum principle. The vector quantity angular momentum It is derived from the expression for a particle's angular momentum.
Given,
mass of ball 1 = m1
m₂ mass of ball 2=m2
v₁ is the velocity of ball=r₁ω₁
v₂ is the velocity of ball 2=r₂ω₂
The total angular momentum is given as;

Hence the total angular momentum will be 
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Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²
Answer:
2.24 m/s
Explanation:
resolving force of 29.2 N in x component
Fx = 29.2 cos 57.7
Fx = 15.6N
as force of friction is 12.7 N hence net force which produces acceleration is
15.6-12.7=2.9 N
by Newton 's law a=f/m
a= 2.9/6.87=0.422 m/s^2
now equation of motion is
v^2= U^2+2as
= 0^2+2(.422)(5.93)
v^2=5.00
v=2.24 m/s